Basics

If a plane intersects a cone, it makes one of the following
0-1 points
0-2 straight lines
Circle
Parabola
Ellipse
Hyperbola
Standard equation for all the things that this intersection makes
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$
Equations to change angle of cartesian plane, assuming B is not 0
$$ x = x\prime\cos(\alpha) - y\prime\sin(\alpha) $$
$$ y = x\prime\sin(\alpha) + y\prime\cos(\alpha) $$
$$ \tan(2\alpha) = \frac{B}{A-C} $$
Eccentricity (e)
the distance to the focus (F) divided by the distance to the directrix (DD')

Circle

$$ (x - h)^2 + (y - k)^2 = r^2 $$
$$ e = 0 $$
$$ a = b $$

Parabola

distance to DD' = distance to f
$$ e = 1 $$
$$ LR = 4p $$
$$ 4p(y - k) = (x - h)^2 $$
It is up (+) or down (-)
$$ 4p(x - h) = (y - k)^2 $$
It is left (-) or right (+)

Ellipse (Paraboloid)

sum of distance to foci = 2a (major axis)
minor axis = 2b
$$ c = \pm\sqrt{a^2 - b^2} $$
$$ e = \frac{distance(f)}{distance(DD\prime)} $$
$$ 0 < e < 1 $$
$$ a > b $$
$$ e = \frac{c}{a} $$
$$ LR = \frac{2b^2}{a} $$
$$ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 $$
It is left-right
$$ a = 2, b = 1, h = k = 0 $$
$$ \frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1 $$
It is up-down
$$ a = 2, b = 1, h = k = 0 $$

Hyperbola

Triangle (?) distance from point on the curve to f, f' = 2a
conjugate ax (?), TA = 2b
Transverse m (?), CA = 2a
Transverse ax = 2a ?
$$ c = \pm\sqrt{a^2 + b^2} $$
$$ e > 1 $$
$$ e = \frac{c}{a} $$
$$ LR = \frac{2b^2}{a} $$
$$ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 $$
It is left-right
$$ a = b = h = k = 0 $$
$$ -\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1 $$
It is up-down
$$ a = b = h = k = 0 $$
Something cool to solidify concept
I can't graph these equations as they are as they are implicitly defined (as in, it isn't y = func(x))
To solve this conundrum, I set $$ y^2 = ... x^2 ...$$
then: $$ y = \sqrt{...x^2...} $$
$$ y = -\sqrt{...x^2...} $$
Which should seem like the familiar $$ y = \pm func(x) $$
And graphed each y equation